My Answer:
R: 54 ; W:9 ; B:75
Pr (R or B) = \(\frac{54+75}{54+9+75}\) = \(\frac{54+75}{54+9+75}\) =0.9348
My Answer:
G: 19 ; R: 20 ; B:24 ; Y: 17
{Total balls= 19+20+24+17 = 80}
Imagice there are 80 spots in a line, and firstly I place a red ball at the end of the line. The rest 79 balls can randomly palce on a line. The numbers of ways to place 4 colors balls can use combination method.
\(\binom {79}{19}\) : From the rest 79 spots, I ramdomly pick 19 spots for Green color balls.
\(\binom {60}{19}\) : From the rest 60 spots, I ramdomly pick 19 spots for the rest of Red color balls.
\(\binom {41}{24}\) : From the rest 41 spots, I ramdomly pick 24 spots for the rest of Blue color balls.
\(\binom {17}{17}\) : From the rest 17 spots are for 17 Yellow balls, actually there is no other option for Yellow spots, it only has one way.
So the number of the way with red ball at the end can be calculated as \(\binom {79}{19} * \binom {60}{19} * \binom {41}{24}\)
For the same process, the number of the way to rearrange 80 balls will be
\(\binom {80}{19} * \binom {61}{20} * \binom {42}{24}\)
Answer :
\(\frac{number of the ways have red at end}{total number of ways in arrange}\)
= \(\frac{\binom {79}{19} * \binom {60}{19} * \binom {41}{24}}{\binom {80}{19} * \binom {61}{20} * \binom {42}{24}}\)
= \(\frac{\binom {79}{{(79-19)}!19!} * \binom {60!}{{(60-19)}!19!} * \binom {41!}{{(41-24)!}24!}}{\binom {80}{{80-19}!19!} * \binom {61!}{{(61-20)}!20!}* \binom {42}{{(42-24)!}24!}}\)
= \(\frac{\frac{79!}{60!19!} * \frac {60!}{41!19!} * \frac{41!}{17!24!}}{\frac {80!}{61!19!} *\frac{61!}{41!20} * \frac{42!}{18!24}}\)
= \(\frac{18*20}{80*42}\)
= 0.1071
| Gender and Residence of Customers | ||
|---|---|---|
| Males | Females | |
| Apartment | 81 | 228 |
| Dorm | 116 | 79 |
| With Parent(s) | 215 | 252 |
| Sorority/Fraternity House | 130 | 97 |
| Other | 129 | 72 |
My Answer:
Since Not man: 228+79+252+97+72=728 since not With parents: 81+116+130+129 = 456
\(Pr(Not man or not With parents)\) =\(\frac{728+456}{1399}\) = 0.8463
My Answer:
Event A : Going to the gym.
Event B : Losing weight.
Event A and B could be dependent if acting is same person.
Event A and B could be independent if acting is not a same person.
My Answer:
\({\binom {8}{3} * \binom {7}{3} * \binom {3}{1}}\)
= \({\frac{8!}{5!3!} * \frac{7!}{4!3!} * \frac{3!}{2!1!}}\)
= 5880
My Answer
Event A: Jeff runs out of gas on the way to work.
Event B: Liz watches the evening news.
Even A and Event B are indepent, since A and B are different people in different actions.
\(\binom {14}{8}\) : Selceting 8 people from the 18 spots \(8!\) : rank matters
My answer= $ * 8! $ = 121080960
My answer: = \(\frac{\binom {4}{1} * \binom {9}{3}} {\binom {22}{4}}\)
= \(\frac{{\frac{4!}{3!1!}} * {\frac{9!}{6!3!}}} {\frac{22!}{18!4!}}\)
= 0.05
= \(\frac{11!}{7!}\) = \(\frac{11*10*9*....*3*2*1}{7*6*5*....*3*2*1}\) = \(11*10*9*8\) = 7920
My Answer:
\(A'\) : 33% of subscribers to a fitness magazine are the age of 34 or younger.
My answer:
Event A : exactly three heads in four tosses Event A’: is complement of A
Pr(A)= \({\frac{\binom {4}{3}}{2^4}}\) = \({\frac{4}{16}}\) = 0.25
=> Pr(A’) = 1 - 0.25 = 0.75
E(X) = \(97* Pr(A)+ (-30.11) * Pr(A')\) = \(97 * 0.25+ (-30) * 0.75\) = 1.75
My Answer:
E(x) = 1.75*559 = 978.25
expect to win 978.25.
My answer:
Event A : 4 tails or less in flipping a coin 9 times
Pr(A)= \(\sum {Pr(0) + Pr(1) +Pr(2) + Pr(3) + Pr(4) }\) = \({\frac{\binom {9}{0}}{2^9}} + {\frac{\binom {9}{1}}{2^9}} + {\frac{\binom {9}{2}}{2^9}} + {\frac{\binom {9}{3}}{2^9}} + {\frac{\binom {9}{4}}{2^9}}\) = \({\frac{1+9+36+84+126}{512}}\) = 0.5
=> Pr(A’) = 1 - 0.5 = 0.5
E(X) = \(23 * Pr(A)+ (-26) * Pr(A')\) = \(23 * 0.5+ (-26) * 0.5\) = -1.5
My Answer:
E(x) = 994 * (-1.5) = -1491
expect to lose 1491.
My Answer:
Since liar is 0.2, then true will be 0.8.
| Infomation | |||
|---|---|---|---|
| ————- | True | liar | |
| Dect True | Real True = 0.72 | False Liar = 0.082 | |
| Dect Liar | False True = 0.08 | Real Liar =0.118 | |
| Total | 0.8 | 0.2 | 1.0 |
Sensitivity :
\({\frac{RL}{RL+FL}}\) = 0.59
Specificity :
\({\frac{RT}{RT+FT}}\)= 0.9
FL+RL = 0.2
RT+FT = 0.8
=> RL = 0.2*0.59 = 0.118
=> FL = 0.2 - 0.118 = 0.082
=> RT = 0.8*0.9 = 0.72
=> RL = 0.8 - 0.72 = 0.08
Pr(liar | D_liar) =\(\frac{Pr(Real.liar\_n\_Dect.liar)}{Pr(Dect_liar)}\) =\(\frac {0.118}{0.118+0.08}\) =0.0.59596
Pr(True | D_True) =\(\frac{Pr(True\_n\_Dect.True)}{Pr(Dect.True)}\) =\(\frac {0.72}{0.72+0.082}\) =0.89776
Pr(liar or D_liar) = Pr(liar) + Pr(Dect_liar) - Pr(liar n Dect_lair) =0.2 + 0.198 - 0.118 =0.28